Mathematics - Definite Integration Question with Solution | TestHub

Let $I={\int }_{-1}^{+1}\frac{\cos \alpha x}{1+3^x}dx$ ...(I) $I={\int }_{-1}^{+1}\frac{\cos \alpha x}{1+3^{-x}}dx$ $\left(u\mathrm{sing}{\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx\right)$ ...(II) Add (1) and (II) $\begin{array}{rl}& 2I={\int }_{-1}^{+1}\cos (\alpha x)dx=2{\int }_0^1\cos (\alpha x)dx\\ & I=\frac{\sin \alpha }{\alpha }=\frac{2}{\pi }(\text{ given })\\ & \therefore \alpha =\frac{\pi }{2}\end{array}$